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 Sequence Group III (Posted on 2008-10-28)
Five positive integers A, B, C, D and E, with A < B < C < D < E, are such that:

(i) A, B and C (in this order) are in arithmetic sequence, and:

(ii) B, C and D (in his order) are in geometric sequence, and:

(iii) C, D and E (in this order) are in harmonic sequence.

Determine the minimum value of (E-A) such that there are precisely two quintuplets (A, B, C, D, E) that satisfy all the given conditions.

Note: Try to solve this problem analytically, although computer program/ spreadsheet solutions are welcome.

 See The Solution Submitted by K Sengupta No Rating

Comments: ( Back to comment list | You must be logged in to post comments.)
 proof--thanks to Dej Mar's formula | Comment 4 of 5 |
(In reply to re: computer solutions -- no proof by Dej Mar)

"Using substitution I was able to derive the equation,
[E-A] = (4/A)*x2 + 4*x,
such that x is the common difference between A and B. "

The above can be used to show that no further solutions exist with E-A at or below 64.

As x is the difference between B and A (the common difference of either B-A or C-B), and that difference must also be less than 64 in order for E-A to be less than or equal to 64, for a solution 4*x^2 must be less than 4*64^2 or 16,384.

But as that is the case, in order for E-A to be an integer, A must be less than 16,384.  Since I checked all the values up through 200,000, all the possibilities are covered for all E-A of less than or equal to 64, and 64 is the answer.

 Posted by Charlie on 2008-10-29 11:58:22

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